Find the equivalent capacitance between the terminals `A` and `B` in the given Fig. Take `C = 1 muF`.

Answer

`0.62 muF

Explanation

Supoose the equivalent capacity between `A` and `B` is `C'`. As the netwrok is infinite, therefore addition or removal of one set of capacitors from the network will nto affect our result. It implies that the capacitance between `D` and `E` can be taken as `C'`. Effective capacitive of parallel combination of `C` and `C'`. Fig is `C" = C + C`

Now `C"` and `C` are in series.
`:. (1)/(C') = (1)/(C ) + (1)/(C") = (1)/(C ) + (1)/(C + C')`
`= (C' + 2C)/(C (C + C'))`
or `C'^(2) + 2C C' = C (C + C') = C^(2) + C C'`
or `C'^(2) + C C' - C^(2) = 0`
`C' = (-C +- sqrt(C^(2) + 4C^(2)))/(2) = (-C +- sqrt(5C^(2)))/(2)`
As capacitance cannot be negative,
`:. C' = (-C + C sqrt(5))/(2) = (C )/(2) [sqrt(5) - 1]`
`C' (C )/(2) (2.24 - 1) = (1.24)/(2) C = 0.62C`
As `C = 1 muF`, therefore,
`C' = 0.62 muF`